## Posts Tagged ‘motor speed’

### Equations Used to Manipulate Torque Relative to Horsepower

Monday, July 28th, 2014
 Last time we learned that we can get more torque out of a motor by using one of two methods.   In the first method we attach a gear train to the motor, then try different gear sizes until we arrive at the desired torque for the application.   In the second method we eliminate the gear train altogether and simply use a higher horsepower motor to give us the torque we need.       Today we’ll explore the second method.   We’ll use the equation presented in our last blog to determine torque, T, relative to a motor’s horsepower, HP, when the motor operates at a speed, n: T = [HP ÷ n] × 63,025       In earlier blogs in this series we employed a gear train attached to an electric motor to power a lathe.   It provided an insufficient 200 inch pounds of torque when 275 is required.   Let’s use the equation and a little algebra to see how much horsepower this motor develops if it turns at a speed of 1750 RPM, a common speed for alternating current (AC) motors: 200 inch pounds = [HP ÷ 1750 RPM] × 63,025 200 = HP × 36.01 HP = 200 ÷ 36.01 = 5.55 horsepower       For the purpose of our example today, let’s say we’ve nonsensically decided not to use a gear train, leaving us with no choice but to replace the underpowered motor with a more powerful one.   So let’s see what size motor we’ll need to provide us with the required horsepower of 275 inch pounds.       Using the torque equation and plugging in numbers already provided our equation becomes: 275 inch pounds = [HP ÷1750 RPM] × 63,025 275 = HP × 36.01 HP = 275 ÷ 36.01 = 7.64 horsepower       This tells us that we need to replace the 5.55 horsepower motor with a 7.65 horsepower motor.       As you might have guessed, the higher the motor’s horsepower, the larger that motor’s size and weight — if you’re not using a gear train, that is.   Bigger, bulkier motors cost more to purchase and operate and also take up more space, which often makes them impractical to use.       All this translates to the reality that sometimes it just makes more sense to use a gear train to provide more torque.   It’s a lot easier and cheaper to attach a gear train to a motor and manipulate its gear sizes to arrive at desired torque than it is to buy a bigger motor.       You may have deduced by now that it’s relatively easy to get more torque.   Almost too easy.   Next time we’ll see how increased torque comes at another type of cost, the cost of speed. _______________________________________

### The Relationship Between Torque and Horsepower

Tuesday, July 15th, 2014
 We’ve been discussing gear trains for some time now, and last time we posed the question:  Why even bother using a gear train and performing complex computations to arrive at a desired torque for an application?   Why not just use a bigger motor to start with?   Today we’ll see why.       First, we must acknowledge that sometimes higher torque is achieved by simply using a more powerful motor.   But sometimes this isn’t possible or practical.       To begin our discussion, we must first understand how torque is related to motor power, the amount of mechanical work a motor can perform.   Torque is in fact a function of how much mechanical power a motor produces.   In the United States motor power is typically measured in units of horsepower.       The following equation illustrates the relationship between torque, horsepower, and motor speed: T = [HP ÷  n]  ×  63,025 where T is the motor shaft’s torque in units of inch-pounds, HP is the motor’s horsepower, and n is the speed of the motor shaft in revolutions per minute (RPM).   The number 63,025 in the equation is a constant used to convert the units of horsepower and RPM into units of torque (inch pounds).   This equation applies to all sources of mechanical power.   Its versatility enables design engineers to easily determine if a mechanical power source can deliver the torque required to drive a particular piece of machinery.       The torque equation above tells us that in order to get a higher torque T for a given speed n, you’ll have to get a motor with a higher HP.   Put another way, if your speed remains constant and you use a motor with higher horsepower, you’ll get more torque for your application simply by increasing the horsepower.       Next time we’ll plug numbers into our equation and see how it all works. _______________________________________

### When Do You Need To Modify Gear Ratio?

Wednesday, February 19th, 2014
 Last time we saw how the involute profile of spur gear teeth ensures smooth contact between gears when they rotate.   Today we’ll see why it’s important to be able to change the rotational speed of the driven gear in relation to that of the driving gear by modifying their gear ratio, the speeds at which gears move relative to one another.       Why would we want to modify the rotational speeds of gears relative to one another?   One reason is to compensate for the fact that alternating electric current (AC) motors drive most modern machinery, and these motors operate at a fixed speed determined by the 60 cycles per second frequency of electricity provided by the utility power grids of North America.   By fixed speed I mean that the motor’s shaft revolves at a single, fixed rate.  It can’t run any faster or slower.   This is fine for some motorized applications, but not others.       Basic machinery such as wood cutting saws, grinders, and blowers function well within the parameters of the AC motor’s fixed speed, because their working parts are intended to rotate at the same rate as the motor’s shaft.   As a matter of fact, in this instance there’s often no need for a gear train, because the working parts can be connected directly to the motor’s shaft, and the machinery will be powered and function correctly.   There are many instances however in which a fixed speed does not match the speed required for more complex machinery to correctly perform precise, specialized tasks.       Take a machine tool meant to cut steel bars, for example.   It has a rotating part meant to cut through the steel during machining, and to properly do so its cutting tool bit must turn at 400 revolutions per minute (RPM).   If it turns any faster, the cut won’t be smooth and the tool bit will overheat and break due to increased friction.   If the AC motor driving the machine tool turns at 1750 RPM, a common speed for such motors, then the tool bit will be turning at a much faster rate than the desired 400 RPM, and this presents a problem.       To solve the problem we need only add a gear train between the motor and the part containing the tool bit, meaning, we must connect the gear train’s driving gear to the motor’s shaft and a driven gear to the part’s shaft.   But in order for this arrangement to work a conversion must take place, that is, we must design the gear train to operate at a specific gear ratio.   By gear ratio, I mean the speeds at which the two gears will rotate relative to one another.       Next time we’ll introduce the gear ratio formulas that make it all work. _______________________________________